Question

If $1,\omega ,{\omega }^2$ are the cube roots of unity, then
$\left|\begin{array}{c}1\\ {\omega }^{2n}\\ {\omega }^n\end{array}\begin{array}{c}{\omega }^n\\ 1\\ {\omega }^{2n}\end{array}\begin{array}{c}{\omega }^{2n}\\ {\omega }^n\\ 1\end{array}\right|$ has value.

• A. $0$
• B. $\omega$
• C. ${\omega }^2$
• D. $\omega +{\omega }^2$

Answer 1

Answer: (A)

$0$

Since, $1,\omega ,\omega +{\omega }^2$ are the cubes of unity.
$\therefore 1+\omega +{\omega }^2=0$
Now, $\left|\begin{array}{c}1\\ {\omega }^{2n}\\ {\omega }^n\end{array}\begin{array}{c}{\omega }^n\\ 1\\ {\omega }^{2n}\end{array}\begin{array}{c}{\omega }^{2n}\\ {\omega }^n\\ 1\end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{c}1+{\omega }^n+{\omega }^{2n}\\ 1+{\omega }^n+{\omega }^{2n}\\ 1+{\omega }^n+{\omega }^{2n}\end{array}\begin{array}{c}{\omega }^n\\ 1\\ {\omega }^{2n}\end{array}\begin{array}{c}{\omega }^{2n}\\ {\omega }^n\\ 1\end{array}\right|$
$=(1+{\omega }^n+{\omega }^{2n})\left|\begin{array}{c}1\\ 1\\ 1\end{array}\begin{array}{c}{\omega }^n\\ 1\\ {\omega }^{2n}\end{array}\begin{array}{c}{\omega }^{2n}\\ {\omega }^n\\ 1\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$=(1+{\omega }^n+{\omega }^{2n})\left|\begin{array}{c}1\\ 0\\ 0\end{array}\begin{array}{c}{\omega }^n\\ 1-{\omega }^n\\ {\omega }^{2n}-{\omega }^n\end{array}\begin{array}{c}{\omega }^{2n}\\ {\omega }^n-{\omega }^{2n}\\ 1-{\omega }^{2n}\end{array}\right|$
$=(1+{\omega }^n+{\omega }^{2n})\{(1-{\omega }^n)(1-{\omega }^{2n})+{({\omega }^n-{\omega }^{2n})}^2\}$
$=(1+{\omega }^n+{\omega }^{2n})(1+{\omega }^{3n}-{\omega }^n-{\omega }^{2n}+{\omega }^{2n}+{\omega }^{4n}-2{\omega }^{3n})$
$=(1+{\omega }^n+{\omega }^{2n})(1+1-2)=0$